An electric field of 20 N C^ -1 is present along the x -axis in space. A charge of -2.0 × 10^ -4 C is transported from point A to point B . Determine the change in electrical potential energy, U_B - U_A , for the cases

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Quantrex Academy · Accepted Answer

Correct answer: C. 0.016 J , 0.008 J , 0.024 J

The change in electrical potential energy is given by the negative of the work done by the electric field.

\Delta U = -W = -q \vec{E} \cdot \Delta \vec{r}

Given

\vec{E} = 20 \hat{i} \text{ N C}^{-1}

and

q = -2.0 \times 10^{-4} \text{ C}

.

\Delta U = -(-2.0 \times 10^{-4}) (20 \hat{i}) \cdot ((x_B - x_A)\hat{i} + (y_B - y_A)\hat{j})

\Delta U = 4.0 \times 10^{-3} (x_B - x_A) \text{ J}

For case (a),

A = (0, 0)

and

B = (4 \text{ m}, 2 \text{ m})

:

\Delta U_a = 4.0 \times 10^{-3} (4 - 0) = 0.016 \text{ J}

For case (b),

A = (4 \text{ m}, 2 \text{ m})

and

B = (6 \text{ m}, 5 \text{ m})

:

\Delta U_b = 4.0 \times 10^{-3} (6 - 4) = 0.008 \text{ J}

For case (c),

A = (0, 0)

and

B = (6 \text{ m}, 5 \text{ m})

:

\Delta U_c = 4.0 \times 10^{-3} (6 - 0) = 0.024 \text{ J}

An electric field of 20 N C^ -1 is present along the x -axis in...

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